CANCELLATIVE FAMILIES OF SETSJames B.. We show how to construct cancellative families of sets with c2 .54797n elements.. Frankl and F¨uredi [2] call such families cancellative.. We ask h
Trang 1CANCELLATIVE FAMILIES OF SETS
James B Shearer IBM Research Division T.J Watson Research Center
P.O Box 218 Yorktown Heights, NY 10598 email: jbs@watson.ibm.com Submitted: March 25, 1996; Accepted: April 20, 1996
Abstract Following [2], we say a family, H, of subsets of a n-element set is
can-cellative if A ∪ B = A ∪ C implies B = C when A, B, C ∈ H We show how to construct
cancellative families of sets with c2 .54797n elements This improves the previous best
bound c2 .52832n and falsifies conjectures of Erd¨os and Katona [3] and Bollobas [1]
AMS Subject Classification 05C65
We will look at families of subsets of a n-set with the property that A ∪B = A∪C ⇒
B = C for any A, B, C in the family Frankl and F¨uredi [2] call such families cancellative
We ask how large cancellative families can be We define f(n) to be the size of the largest possible cancellative family of subsets of a n-set and f (k, n) to be the size of the largest possible cancellative family of k-subsets of a n-set.
Note the condition A ∪ B = A ∪ C ⇒ B = C is the same as the condition B4C ⊆
A ⇒ B = C where 4 denotes the symmetric difference.
Let F1 be a family of subsets of a n1-set, S1 Let F2 be a family of subsets of a n2
-set, S2 We define the product F1× F2 to be the family of subsets of the (n1+ n2)-set,
S1∪ S2 , whose members consist of the union of any element of F1 with any element of
F2
It is easy to see that the product of two cancellative families is also a cancellative
family ((A1, A2)∪ (B1 , B2) = (A1, A2)∪(C1 , C2)⇒ (A1 ∪ B1 , A2∪ B2 ) = (A1∪ C1 , A2∪ C2) ⇒ A1 ∪ B1 = A1 ∪ C1 and A2 ∪ B2 = A2 ∪ C2 ⇒ B1 = C1 and B2 = C2 ⇒
(B1, B2) = (C1, C2)) Hence f (n1 + n2) ≥ f(n1 )f(n2) Similarly f (k1+ k2, n1+ n2)≥
f (k1, n1 )f (k2, n2)
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1
Trang 2It is easy to show that f (n1+n2)≥ f(n1 )f (n2) implies that limn →∞ n1lg(f (n)) exists
(lg means log base 2) Let this limit be α Note that α ≥ 1
n lg (f (n)) for any fixed n.
Clearly f (1, n) = n as we may take all the 1-element sets Let Hn be the family
of all 1-element sets of a n-set It had been conjectured that the largest cancellative families could be built up by taking products of the families Hn For example Bollobas
conjectured [1] that
f (k, n) =
k
Y
i=1
which comes from letting n = n1+· · · + n k where the ni are as nearly equal as possible
and considering the family H n1×· · ·×H n k When k = 2 determining f (2, n) is the same
as determining how many edges a triangle-free graph can contain So in this case (1)
follows from Turan’s theorem Bollobas [1] proved (1) for k = 3 Sidorenko [4] proved (1) when k = 4 Frankl and F¨ uredi [2] proved (1) for n ≤ 2k However, we will show
below that (1) is false in general
Also Erd¨os and Katona conjectured (see [3]) that (for n > 1) the families achieving
f (n) could be built up as products of H3 and H2 taking as many H3’s as possible So for example
This would mean α = lg33 = 52832+ However, as we will see this conjecture is false as well In fact we show α ≥ 54797+.
We now describe the construction which is the main result of this paper Fix m ≥ 3.
Chose m − 1 integers n1, , n m −1 from {0, 1, 2} so that n1+· · · + n m −1 ≡ 0 mod 3.
Chose an integer h from {1, , m} Clearly these choices can be made in m3 m −2 ways.
We now form a cancellative family of subsets of a 3m-set containing m3 m−2 elements as
follows Identify subsets of a 3m-set with 0,1 vectors of length 3m in the usual way Let the 3m vectors consist of m subvectors of length 3 Let v0 = (100), v1 = (010), v2 = (001)
and w = (111) Form a 3m-vector from our choices above as follows Let the hth 3-subvector be w Let the remaining m − 1 3-subvectors be v n1, , v n m −1 in order Let
F be the family consisting of all 3m-vectors we can form in this way Clearly each of the m3 m −2 choices gives a different vector so F contains m3 m −2 elements We claim F is
a cancellative family For let B, C be two different vectors in F and look at B 4C We
claim B 4C contains at least two 3-subvectors with two 1’s There are two cases If the
3-subvector w is in different positions in B and C then the 3-subvectors in B 4C in these
positions contain two 1’s Alternatively, if the 3-subvector w is in the same position in
B and C then the condition n1+· · ·+n m−1 ≡ 0 mod 3 insures that at least two of the n i differ between B and C (assuming B and C are distinct) and the 3-subvectors in these positions of B 4C contain two 1’s However, this means B4C ⊆ A ∈ F is impossible
(unless B = C) because all elements of F contain only one 3-subvector containing two
or more 1’s
Hence we have
Trang 3f(m + 2, 3m) ≥ m3 m−2 (4)
Clearly (3) is better than (2) for m > 9 We also have α ≥ 1
3m lg(m3 m−2) This is
maximized for m = 24 giving α ≥ 54797+ So we have counter examples to the Erd¨os
and Katona conjecture
Furthermore (4) is better than (1) for m ≥ 8 So the Bollobas conjecture fails for
k ≥ 10.
The idea of the above construction which improves on products of H3 can be applied
to products of other families as well For example, we can do better than (1) starting
with products of Hk for any k > 3 as well Or we can start with the families F
constructed above This will allow a very slight improvement in the lower bound found
for α above.
The best upper bound known for α, α < lg(3/2) = 58496+, is due to Frankl and
F¨uredi [2]
The author thanks Don Coppersmith for bringing this problem to his attention
References
[1] B Bollob´as, Three-Graphs without two triples whose symmetric difference is contained in a third,
Discrete Mathematics 8 (1974), 21–24.
[2] P Frankl and Z F¨uredi, Union-free Hypergraphs and Probability Theory, European Journal of
Combinatorics 5 (1984), 127–131.
[3] G.O.H Katona, Extremal Problems for Hypergraphs, Combinatorics, Mathematical Centre Tracts
part 2 56 (1974), 13–42.
[4] A.F Sidorenko, The Maximal Number of Edges in a Homogeneous Hypergraph containing no
pro-hibited subgraphs, Mathematical Notes 41 (1987), 247-259 (translation from Mathematicheskie
Za-metki 41 (1987), 433-455).